32 bit single precision IEEE 754 binary floating point number 1 - 1000 0110 - 000 0011 0000 0000 0000 0000 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point:
1 - 1000 0110 - 000 0011 0000 0000 0000 0000
to decimal system (base ten)

1. Identify the elements that make up the binary representation of the number:

First bit (the leftmost) indicates the sign,
1 = negative, 0 = positive.


The next 8 bits contain the exponent:
1000 0110


The last 23 bits contain the mantissa:
000 0011 0000 0000 0000 0000

2. Convert the exponent from binary (base 2) to decimal (base 10):

The exponent is allways a positive integer.

1000 0110(2) =


1 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 0 × 20 =


128 + 0 + 0 + 0 + 0 + 4 + 2 + 0 =


128 + 4 + 2 =


134(10)

3. Adjust the exponent.

Subtract the excess bits: 2(8 - 1) - 1 = 127, that is due to the 8 bit excess/bias notation:

Exponent adjusted = 134 - 127 = 7


4. Convert the mantissa from binary (base 2) to decimal (base 10):

Mantissa represents the number's fractional part (the excess beyond the number's integer part, comma delimited)

000 0011 0000 0000 0000 0000(2) =

0 × 2-1 + 0 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 1 × 2-6 + 1 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 0 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 2-14 + 0 × 2-15 + 0 × 2-16 + 0 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 0 × 2-22 + 0 × 2-23 =


0 + 0 + 0 + 0 + 0 + 0.015 625 + 0.007 812 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0.015 625 + 0.007 812 5 =


0.023 437 5(10)

5. Put all the numbers into expression to calculate the single precision floating point decimal value:

(-1)Sign × (1 + Mantissa) × 2(Exponent adjusted) =


(-1)1 × (1 + 0.023 437 5) × 27 =


-1.023 437 5 × 27 =


-131

Conclusion:

1 - 1000 0110 - 000 0011 0000 0000 0000 0000
converted from
32 bit single precision IEEE 754 binary floating point
to
base ten decimal system (float) =

-131(10)

More operations of this kind:

1 - 1000 0110 - 000 0010 1111 1111 1111 1111 = ?

1 - 1000 0110 - 000 0011 0000 0000 0000 0001 = ?


Convert 32 bit single precision IEEE 754 floating point standard binary numbers to base ten decimal system (float)

32 bit single precision IEEE 754 binary floating point standard representation of numbers requires three building blocks: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits), mantissa (23 bits)

Latest 32 bit single precision IEEE 754 floating point binary standard numbers converted to decimal base ten (float)

1 - 1000 0110 - 000 0011 0000 0000 0000 0000 = ? Jan 24 20:48 UTC (GMT)
0 - 1000 0100 - 000 0101 0101 0100 1111 1101 = ? Jan 24 20:48 UTC (GMT)
1 - 1000 0110 - 100 1000 0000 0111 0000 0000 = ? Jan 24 20:47 UTC (GMT)
1 - 1000 0010 - 011 1011 0000 0000 0000 0100 = ? Jan 24 20:47 UTC (GMT)
0 - 1000 0101 - 111 0110 1110 0110 0110 1000 = ? Jan 24 20:47 UTC (GMT)
1 - 1000 1101 - 010 0000 1100 1100 1101 1000 = ? Jan 24 20:47 UTC (GMT)
0 - 0000 0000 - 101 1111 1100 0000 0000 0000 = ? Jan 24 20:47 UTC (GMT)
1 - 1000 0100 - 101 1010 0111 1101 0011 1001 = ? Jan 24 20:46 UTC (GMT)
0 - 1001 1000 - 011 1110 0101 1010 0111 0101 = ? Jan 24 20:46 UTC (GMT)
0 - 1000 0000 - 001 0101 1100 0010 1000 1111 = ? Jan 24 20:46 UTC (GMT)
0 - 1011 1100 - 100 0010 1011 0110 1011 1111 = ? Jan 24 20:45 UTC (GMT)
1 - 1100 0011 - 100 0100 0000 0000 0000 0010 = ? Jan 24 20:45 UTC (GMT)
0 - 0011 1011 - 101 0000 0000 0000 0000 0000 = ? Jan 24 20:44 UTC (GMT)
All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

How to convert numbers from 32 bit single precision IEEE 754 binary floating point standard to decimal system in base 10

Follow the steps below to convert a number from 32 bit single precision IEEE 754 binary floating point representation to base 10 decimal system:

  • 1. Identify the three elements that make up the binary representation of the number:
    First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.
    The next 8 bits contain the exponent.
    The last 23 bits contain the mantissa.
  • 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10).
  • 3. Adjust the exponent, subtract the excess bits, 2(8 - 1) - 1 = 127, that is due to the 8 bit excess/bias notation.
  • 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10).
  • 5. Put all the numbers into expression to calculate the single precision floating point decimal value:
    (-1)Sign × (1 + Mantissa) × 2(Exponent adjusted)

Example: convert the number 1 - 1000 0001 - 100 0001 0000 0010 0000 0000 from 32 bit single precision IEEE 754 binary floating point system to base 10 decimal system (float):

  • 1. Identify the elements that make up the binary representation of the number:
    First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.
    The next 8 bits contain the exponent: 1000 0001
    The last 23 bits contain the mantissa: 100 0001 0000 0010 0000 0000
  • 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10):
    1000 0001(2) =
    1 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20 =
    128 + 0 + 0 + 0 + 0 + 0 + 0 + 1 =
    128 + 1 =
    129(10)
  • 3. Adjust the exponent, subtract the excess bits, 2(8 - 1) - 1 = 127, that is due to the 8 bit excess/bias notation:
    Exponent adjusted = 129 - 127 = 2
  • 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10):
    100 0001 0000 0010 0000 0000(2) =
    1 × 2-1 + 0 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 0 × 2-6 + 1 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 0 × 2-11 + 0 × 2-12 + 0 × 2-13 + 1 × 2-14 + 0 × 2-15 + 0 × 2-16 + 0 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 0 × 2-22 + 0 × 2-23 =
    0.5 + 0 + 0 + 0 + 0 + 0 + 0.007 812 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 061 035 156 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =
    0.5 + 0.007 812 5 + 0.000 061 035 156 25 =
    0.507 873 535 156 25(10)
  • 5. Put all the numbers into expression to calculate the single precision floating point decimal value:
    (-1)Sign × (1 + Mantissa) × 2(Exponent adjusted) =
    (-1)1 × (1 + 0.507 873 535 156 25) × 22 =
    -1.507 873 535 156 25 × 22 =
    -6.031 494 140 625
  • 1 - 1000 0001 - 100 0001 0000 0010 0000 0000 converted from 32 bit single precision IEEE 754 binary floating point representation to decimal number (float) in decimal system (in base 10) = -6.031 494 140 625(10)